To better understand certain problems involving aircraft and propulsion it is necessary to use some mathematical ideas from trigonometry the study of triangles. Let us begin with some definitions and terminology which we will use on this slide. We start with a right triangle. A right triangle is a three sided figure with one angle equal to 90 degrees. A 90 degree angle is called a right angle and that is where the right triangle gets its name. We define the side of the triangle opposite from the right angle to be the hypotenuse, h. It is the longest side of the three sides of the right triangle. The word "hypotenuse" comes from two Greek words meaning "to stretch", since this is the longest side. We are going to label the other two sides a and b. The Pythagorean Theorem is a statement relating the lengths of the sides of any right triangle.

The theorem states that:

Mathematically, this is written:

The theorem has been known in many cultures, by many names, for many years. Pythagoras, for whom the theorem is named, lived in ancient Greece, 2500 years ago. It is believed that he learned the theorem during his studies in Egypt. The Egyptians probably knew of the relationship for a thousand years before Pythagoras. The Egyptians knew of this relationship for a triangle with sides in the ratio of "3 - 4 - 5".

Pythagoras generalized the result to any right triangle. There are many different algebraic and geometric proofs of the theorem. Most of these begin with a construction of squares on a sketch of a basic right triangle. On the figure at the top of this page, we show squares drawn on the three sides of the triangle. A square is the special case of a rectangle in which all the sides are equal in length. The area A of a rectangle is the product of the sides. So for a square with a side equal to a, the area is given by:

So the Pythagorean theorem states the area h^2 of the square drawn on the hypotenuse is equal to the area a^2 of the square drawn on side a plus the area b^2 of the square drawn on side b.

Here's an interactive Java program that let's you see that this area relationship is true:

We begin with a right triangle on which we have constructed squares on the two sides, one red and one blue. We are going to break up the pieces of these two squares and move them into the grey square area on the hypotenuse. We won't loose any material during the operation. So if we can exactly fill up the square on the hypotenuse, we have shown that the areas are equal. You work through the construction by clicking on the button labeled "Next". You can go "Back" and repeat a section, or go all the way back to the beginning by clicking on "Reset".

What is it doing? The first step rotates the triangle down onto the blue square. This cuts the blue square into three pieces, two triangles and a red rectangle. The two triangles are exactly the same size as the original triangle. The "bottom" of the original triangle exactly fits the vertical side of the square, because the sides of a square are equal. The red rectangle has its vertical sides equal to the base of the original triangle, and its horizontal sides equal to the difference between the "bottom" side and the "vertical" side of the original triangle. Using the terminology from the figure at the top of this page, the dimensions of the red rectangle are:

The next step is to move the red rectangle over adjacent to the red square. The rectangle sticks out the top of the red square and the two triangles remain in the blue square. The next step is to move one of the blue triangles vertically into the hypotenuse square. It fits exactly along the side of the hypotenuse square because the sides of a square are equal. The next step is to move the other blue triangle into the hypotenuse square. (We are half way there!) The next step is to slide the form of the original triangle to the left into the red region. The triangle cuts the red region into three pieces, two triangles and a small yellow square. The original triangle fits exactly into this region because of two reasons; the vertical sides are identical, and the horizontal side of the red region is equal to the length of the red square plus the horizontal length of the red rectangle which we moved. The horizontal length of the red region is: